OFFSET
1,3
COMMENTS
Permutations which contain a cycle with length >=3.
For elements of order >= 3 we can show that they can be represented as a product of two involutions. [Proof: Write down the reduced cycle representation of the element. Decompose each sub-cycle of order >=3 into a product of two involutions, and distribute/merge the 1 or 2 factors of the involutions. For sub-cycles of order o>=3 it suffices to show that the reduced cycle (1 2 3 4 ... o), equivalent to the 1-line representation (2 3 4 .. o 1), can be written as a product of 2 involutions, because other cycles are mapped to such a product by a fixed permutation. A constructive proof here is: the reduced cycle (1 2 3 4 ... o) is the product of the two involutions (1 2)(3 o)(4 o-1)(5 o-2)... and (2 o)(3 o-1)... The first involution swaps the first two elements and reverts the ordering of all others from the 3rd on, then the second involution reverts the order of the elements from the 2nd on. The net effect is a cyclic shift of the o elements.] - R. J. Mathar, Jan 04 2017
a(n) is the number of permutations of [n] that are not involutions (see first formula). - Joerg Arndt, Jul 06 2020
FORMULA
a(n) = n! - A000085(n).
E.g.f.: 1/(1-x) - exp(x*(x+2)/2).
Conjecture: (-n+3)*a(n) +(n^2-2*n-1)*a(n-1) +(-n+1)*a(n-2) -(n-1)*(n-2)^2*a(n-3)=0. - R. J. Mathar, Jan 04 2017
Conjecture: a(n) +(-n-3)*a(n-1) +(2*n-1)*a(n-2) +n*(n-2)*a(n-3) -2*(n-2)*(n-3)*a(n-4)=0. - R. J. Mathar, Jan 04 2017
MATHEMATICA
nn=20; Range[0, nn]! CoefficientList[Series[1/(1-x)-Exp[x+x^2/2], {x, 0, nn}], x] (* Geoffrey Critzer, Jan 09 2013 *)
CROSSREFS
KEYWORD
nonn,easy
AUTHOR
Sharon Sela (sharonsela(AT)hotmail.com), Dec 29 2001
EXTENSIONS
More terms from Vladeta Jovovic, Dec 31 2001
STATUS
approved